3x3 system

[logistic_guy]

here is the question

Solve $\displaystyle \bold{X}' = \begin{bmatrix}1 & -2 & 2 \\-2 & 1 & -2 \\2 & -2 & 1 \end{bmatrix} \bold{X}$.


my attemb
$\displaystyle \bold{Z} = \bold{A} - \lambda\bold{I} = \begin{bmatrix}1 & -2 & 2 \\-2 & 1 & -2 \\2 & -2 & 1 \end{bmatrix} - \lambda \begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \end{bmatrix} = \begin{bmatrix}1 & -2 & 2 \\-2 & 1 & -2 \\2 & -2 & 1 \end{bmatrix} + \begin{bmatrix}-\lambda & 0 & 0 \\0 & -\lambda & 0 \\0 & 0 & -\lambda \end{bmatrix} = \begin{bmatrix}1 - \lambda & -2 & 2 \\-2 & 1 - \lambda & -2 \\2 & -2 & 1 - \lambda \end{bmatrix}$

det$\displaystyle (\bold{Z}) = (1 - \lambda)[(1 - \lambda)(1 - \lambda) - (-2)(-2)] - -2[(-2)(1 - \lambda) - (-2)(2)] + 2[(-2)(-2) - (1 - \lambda)(2)]$

$\displaystyle = (1 - \lambda)[1 - \lambda - \lambda + \lambda^2 - 4] + 2[-2 + 2\lambda + 4] + 2[4 - 2 + 2\lambda]$

$\displaystyle = (1 - \lambda)[-2\lambda + \lambda^2 - 3] + 2[2\lambda + 2] + 2[2 + 2\lambda]$

$\displaystyle = -2\lambda + \lambda^2 - 3 + 2\lambda^2 - \lambda^3 + 3\lambda + 4\lambda + 4 + 4 + 4\lambda$

$\displaystyle = 9\lambda + 3\lambda^2 - \lambda^3 + 5 = 0$

this mean

$\displaystyle \lambda^3 - 3\lambda^2 - 9\lambda - 5 = 0$

i'm strugle to solve degree $\displaystyle 3$ so i learn rational root theorem

this give me possible rational roots $\displaystyle \pm 1$ and $\displaystyle \pm 5$

testing one by one

$\displaystyle \lambda(1) = 1^3 - 3(1)^2 - 9(1) - 5 = 1 - 3 - 9 - 5 = -16 \neq 0$

$\displaystyle \lambda(-1) = (-1)^3 - 3(-1)^2 - 9(-1) - 5 = -1 - 3 + 9 - 5 = 0$
so $\displaystyle (\lambda + 1)$ is factor

$\displaystyle \frac{\lambda^3 - 3\lambda^2 - 9\lambda - 5}{\lambda + 1} = \lambda^2 - 4\lambda - 5$ i don't know how to write long division steps in latex

i've get $\displaystyle (\lambda + 1)(\lambda^2 - 4\lambda - 5) = (\lambda + 1)(\lambda + 1)(\lambda - 5) = (\lambda + 1)^2(\lambda - 5) = 0$

this mean $\displaystyle \lambda_1 = \lambda_2 = -1$ and $\displaystyle \lambda_3 = 5$

i get the same problem as before$\displaystyle \bold{X_1} = \bold{X_2}$

this give me only two solutions $\displaystyle \bold{X_1}$ and $\displaystyle \bold{X_3}$but i need three solutions

 

[topsquark]

Did you look up what you need to do if you have repeated roots yet?

-Dan

 

[logistic_guy]

thank

Did you look up what you need to do if you have repeated roots yet?

-Dan

this step i don't know
can you show me how to do it?

 

[Dr.Peterson]

this step i don't know
can you show me how to do it?

Did you look up what you need to do if you have repeated roots yet?

You don't know how to search (in your textbook or online)? Please try.

I searched with "linear differential equation repeated eigenvalue". Here's one of many sources I found:

Differential Equations - Repeated Eigenvalues

In this section we will solve systems of two linear differential equations in which the eigenvalues are real repeated (double in this case) numbers. This will include deriving a second linearly independent solution that we will need to form the general solution to the system. We will also show...

tutorial.math.lamar.edu

 

[logistic_guy]

You don't know how to search (in your textbook or online)? Please try.

I searched with "linear differential equation repeated eigenvalue". Here's one of many sources I found:

Differential Equations - Repeated Eigenvalues

In this section we will solve systems of two linear differential equations in which the eigenvalues are real repeated (double in this case) numbers. This will include deriving a second linearly independent solution that we will need to form the general solution to the system. We will also show...

tutorial.math.lamar.edu

the notation is very complicated
can you explain to me the idea?

 

[topsquark]

the notation is very complicated
can you explain to me the idea?

You passed this class, did you not? This is review!

-Dan